Smooth quotients of bi-elliptic surfaces

We consider the quotient X of bi-elliptic surface by a finite automorphism group. If X is smooth, then it is a bi-elliptic surface or ruled surface with irregularity one. As a corollary any bi-elliptic surface cannot be Galois covering of projective plane, hence does not have any Galois embedding.


Statement of result
We consider a covering of surface, i.e., let X 1 and X 2 be connected normal complex surfaces and π : X 1 −→ X 2 a finite surjective proper holomorphic map. There are lots of studies of the covering. We are interested in the following cases (1) and (2): (1) X 2 is a rational surface, in particular the projective plane P 2 , for example (Friedman and Leyenson 2010) (2) X 1 has Kodaira dimension 0, for example (Muammed Uludaǧ 2005;Yoshihara 2007) In this note we consider the case where X 1 is a bi-elliptic surface and π is a Galois covering. The definition of bi-elliptic surface is as follows, which has been called a hyperelliptic surface (Suwa 1970 Definition 1 A bi-elliptic surface is a surface with the geometric genus zero and having an abelian surface as its unramified covering.
First we note the following: Remark 2 Let S be a bi-elliptic surface. If π : S −→ X is a Galois covering and X is smooth, then X has no curve with negative self-intersection number. In particular, if X is rational, then it is P 2 or P 1 × P 1 .
Proof Suppose the contrary. Then, there exists an irreducible curve C in X such that C 2 < 0. We have that (π * (C)) 2 = deg π · C 2 < 0. Thus there exists an irreducible component C in π * (C) with C 2 < 0. which is a contradiction. Because there exists no curve with negative self-intersection number in bi-elliptic surface.
Our result is stated as follows: Theorem 3 Let S be a bi-elliptic surface and G a finite subgroup of Aut(S), the automorphism group of S. Let X = S/G be the quotient surface. If X is smooth, then it is a bi-elliptic surface or a ruled surface with irregularity one.
As a corollary we have the following.
Corollary 4 Bi-elliptic surface cannot be a Galois covering of any smooth rational surface. Miranda (2005) presents a construction of bi-elliptic surface as covering of rational elliptic surface. We present the other examples.
Example 5 Let A be the abelian surface with the period matrix where λ > 0 and e n = exp(2π √ −1/n), (n = 4, 6). Let σ i (i = 1, 2) be the automorphism of A induced by where z ∈ C 2 . Then S = A/ σ m 1 (n = 2m) and X = A/ σ 1 are bi-elliptic surfaces. Note that π : S −→ X is a double covering. On the other hand, we have σ m 1 σ 2 = σ 2 σ m 1 , hence σ 2 induces an automorphism σ 2 on S. It is easy to see that S/ σ 2 is isomorphic to E × P 1 , where E is an elliptic curve with the period matrix (1/2, λ).
Remark 6 [Yoshihara (2000), Example 2.4] Let A i (i = 1, 2) be the abelian surface defined by the following period matrix: we see that τ i defines an automorphism of S i and S i /τ i turns out to be a rational surface. Note that the rational surfaces in these examples have singular points.
We have defined Galois embedding of algebraic variety and applied it to study several varieties. In particular we have shown that many abelian surfaces have Galois embeddings (Yoshihara 2007). However, as a corollary of Theorem 3, we have the following.
Corollary 7 There does not exist any Galois embedding of bi-elliptic surface.

Proof
Let S be a bi-elliptic surface and π 1 : S −→ X a finite Galois covering. Then S can be expressed as C 2 / 1 , where 1 is the complex crystallographic group given in [Suwa (1970), Theorem]. Let A = C 2 / 0 be the abelian surface, where π 2 : A −→ S is an unramified covering such that 0 is the normal subgroup of 1 and | 1 : 0 | = 2, 3, 4 or 6. Put n = | 1 : 0 |. Let G be the finite subgroup of Aut(S) such that X = S/G. For g ∈ G, we let g be the lift of g on the universal covering C 2 .

Claim 8 The g turns out to be an affine transformation.
Proof For each s ∈ n 0 , there exists t ∈ 0 such that s = t n . Since 0 ⊂ 1 and 1 is a discrete group, we have u ∈ 1 such that gt = u g. Hence we get gt n = u n g and u n ∈ 0 . This implies gs ≡ g(mod 0 ). Let L 0 be the lattice defining the abelian surface A = C 2 / 0 . Putting z = t (z 1 , z 2 ) ∈ C 2 , g(z 1 , z 2 ) = ( g 1 (z 1 , z 2 ), g 2 (z 1 , z 2 )), we have where i = 1, 2, t (α 1 , α 2 ) ∈ nL 0 and t (β 1 , β 2 ) ∈ L 0 . Hence we get This means ∂ g i /∂z j is a holomorphic function on C 2 / n 0 , which is also an abelian surface. Therefore ∂ g i /∂z j is a constant, i.e., g is an affine transformation.
Thus g has the representation gz = M(g)z + v(g), where z ∈ C 2 , M(g) ∈ G L(2, C) and v(g) ∈ C 2 . We use this expression hereafter. Let 2 be the affine transformation group generated by { g | g ∈ G} and 1 . Then we have X = C 2 / 2 . Since π 1 : S −→ X is a Galois covering, we have 0 (resp. 1 ) is a normal subgroup of 1 (resp. 2 ) and 2 / 1 ∼ = G. Let 1 / 0 be generated by σ . Then, we have the 1 0 0 e n and v(σ ) = 1/n 0 Claim 9 The X has two fibrations.
Proof Since 1 is a normal subgroup of 2 , there exists an integer r such that g σ g −1 = t σ r , where g ∈ G and t ∈ 0 . This means M(g)M(σ )M(g) −1 = M(σ ) r , hence is easy to see that M(g) is a diagonal matrix. Hence the lift of each element γ of 2 can be expressed as Since L 0 is the lattice defining the abelian surface A, Referring to the list of abelian surfaces in Suwa (1970), we infer that C/ 0i is an elliptic curve. Using the above representation ( * ), we define Then 0i is a subgroup of 2i with a finite index. Hence 0i is a discrete subgroup and C/ 2i is a smooth curve. Thus we have the following diagram Suppose that X is smooth. Then we have and the Kodaira dimension of X is less than or equal to 0. By Remark 2 and the classification theorem of algebraic surfaces, we infer that X is P 1 × P 1 , a ruled surface with irregularity one or a bi-elliptic surface.
Claim 10 The X cannot be P 1 × P 1 .
Proof In the representation above ( * ) there exists i such that α i = 1. Then the fiber spacep i : X −→ C i has a multiple singular fiber. Let F i be the fiber of the projection π i : P 1 × P 1 −→ P 1 . Then each divisor D is linearly equivalent to n 1 F 1 + n 2 F 2 . If D 2 = 0, then we have n 1 = 0 or n 2 = 0. So that there are only two fibrations from P 1 × P 1 to P 1 , which are the first and second projections. Neither of projections have any multiple fiber. Combining the above assertions we infer readily the conclusion.
This completes the proof.
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